Chapter 04

January 27, 2005 11:44 L24-ch04 Sheet number 1 Page number 127 black

CHAPTER 4
Derivatives of Logarithmic, Exponential, and
Inverse Trigonometric Functions

EXERCISE SET 4.1
2
1. y = (2x − 5)1/3 ; dy/dx = (2x − 5)−2/3
3
1 −2/3 2 −2/3
2. dy/dx = 2 + tan(x2 ) sec2 (x2 )(2x) = x sec2 (x2 ) 2 + tan(x2 )
3 3
−1/3
2 x+1 x − 2 − (x + 1) 2
3. dy/dx = =−
3 x−2 (x − 2)2 (x + 1)1/3 (x − 2)5/3

−1/2 −1/2
1 x2 + 1 d x2 + 1 1 x2 + 1 −12x 6x
4. dy/dx = = =− √
2 x2 − 5 dx x2−5 2 x2 − 5 (x2 − 5)2 (x2 − 5)3/2 x2 + 1

2 1 2
5. dy/dx = x3 − (5x2 + 1)−5/3 (10x) + 3x2 (5x2 + 1)−2/3 = x (5x2 + 1)−5/3 (25x2 + 9)
3 3
√
3
2x − 1 1 2 −4x + 3
6. dy/dx = − + = 2
x2 x 3(2x − 1)2/3 3x (2x − 1)2/3

5 15[sin(3/x)]3/2 cos(3/x)
7. dy/dx = [sin(3/x)]3/2 [cos(3/x)](−3/x2 ) = −
2 2x2
1 −3/2 3 2 −3/2
8. dy/dx = − cos(x3 ) − sin(x3 ) (3x2 ) = x sin(x3 ) cos(x3 )
2 2

dy dy 6x2 − y − 1
9. (a) 1 + y + x − 6x2 = 0, =
dx dx x
2 + 2x3 − x 2 dy 2
(b) y = = + 2x2 − 1, = − 2 + 4x
x x dx x
dy 1 1 1 1 2 2
(c) From Part (a), = 6x − − y = 6x − − + 2x2 − 1 = 4x −
dx x x x x x x2

1 −1/2 dy dy √
10. (a) y − cos x = 0 or = 2 y cos x
2 dx dx
dy
(b) y = (2 + sin x) = 4 + 4 sin x + sin2 x so
2
= 4 cos x + 2 sin x cos x
dx
dy √
(c) from Part (a), = 2 y cos x = 2 cos x(2 + sin x) = 4 cos x + 2 sin x cos x
dx
dy dy x
11. 2x + 2y = 0 so =−
dx dx y

dy dy dy 3y 2 − 3x y 2 − x2
12. 3x2 + 3y 2 = 3y 2 + 6xy , = 2 = 2
dx dx dx 3y − 6xy y − 2xy

dy dy
13. x2 + 2xy + 3x(3y 2 ) + 3y 3 − 1 = 0
dx dx
dy dy 1 − 2xy − 3y 3
(x2 + 9xy 2 ) = 1 − 2xy − 3y 3 so =
dx dx x2 + 9xy 2

127


128 Chapter 4

dy dy
14. x3 (2y) + 3x2 y 2 − 5x2 − 10xy + 1 = 0
dx dx
dy dy 10xy − 3x2 y 2 − 1
(2x3 y − 5x2 ) = 10xy − 3x2 y 2 − 1 so =
dx dx 2x3 y − 5x2
dy
1 dy y 3/2
15. − − dx = 0, = − 3/2
2x3/2 2y 3/2 dx x

(x − y)(1 + dy/dx) − (x + y)(1 − dy/dx)
16. 2x = ,
(x − y)2
dy dy x(x − y)2 + y
2x(x − y)2 = −2y + 2x so =
dx dx x

dy dy 1 − 2xy 2 cos(x2 y 2 )
17. cos(x2 y 2 ) x2 (2y) + 2xy 2 = 1, =
dx dx 2x2 y cos(x2 y 2 )

dy dy dy y 2 sin(xy 2 )
18. − sin(xy 2 ) y 2 + 2xy = , =−
dx dx dx 2xy sin(xy 2 ) + 1

dy dy
19. 3 tan2 (xy 2 + y) sec2 (xy 2 + y) 2xy + y2 + =1
dx dx
dy 1 − 3y 2 tan2 (xy 2 + y) sec2 (xy 2 + y)
so =
dx 3(2xy + 1) tan2 (xy 2 + y) sec2 (xy 2 + y)

(1 + sec y)[3xy 2 (dy/dx) + y 3 ] − xy 3 (sec y tan y)(dy/dx) dy
20. = 4y 3 ,
(1 + sec y)2 dx
dy
multiply through by (1 + sec y)2 and solve for to get
dx
dy y(1 + sec y)
=
dx 4y(1 + sec y)2 − 3x(1 + sec y) + xy sec y tan y

2
dy dy 2x dy d2 y
21. 4x − 6y = 0, = , 4−6 − 6y = 0,
dx dx 3y dx dx2
2
d2 y 3 dy
dx −2 2(3y 2 − 2x2 ) 8
=− = =− 3
dx2 3y 9y 3 9y

dy x2 d2 y y 2 (2x) − x2 (2ydy/dx) 2xy 2 − 2x2 y(−x2 /y 2 ) 2x(y 3 + x3 )
22. = − 2, =− =− =− ,
dx y dx2 y4 y4 y5
d2 y 2x
but x3 + y 3 = 1 so =− 5
dx2 y

dy y d2 y x(dy/dx) − y(1) x(−y/x) − y 2y
23. =− , =− =− = 2
dx x dx2 x2 x2 x

2
dy dy dy y dy d2 y dy d2 y d2 y 2y(x + y)
24. y + x + 2y = 0, =− ,2 +x 2 +2 + 2y 2
= 0, =
dx dx dx x + 2y dx dx dx dx dx2 (x + 2y)3

dy d2 y dy sin y
25. = (1 + cos y)−1 , = −(1 + cos y)−2 (− sin y) =
dx dx2 dx (1 + cos y)3


Exercise Set 4.1 129

dy cos y
26. = ,
dx 1 + x sin y
d2 y (1 + x sin y)(− sin y)(dy/dx) − (cos y)[(x cos y)(dy/dx) + sin y]
=
dx2 (1 + x sin y)2
2 sin y cos y + (x cos y)(2 sin2 y + cos2 y)
=− ,
(1 + x sin y)3
but x cos y = y, 2 sin y cos y = sin 2y, and sin2 y + cos2 y = 1 so
d2 y sin 2y + y(sin2 y + 1)
=−
dx2 (1 + x sin y)3

dy x √ dy √
27. By implicit diﬀerentiation, 2x + 2y(dy/dx) = 0, = − ; at (1/2, 3/2), = − 3/3; at
dx y dx
√ dy √ √ dy −x
(1/2, − 3/2), = + 3/3. Directly, at the upper point y = 1 − x 2, = √ =
dx dx 1 − x2
1/2 √ √ dy x √
− = −1/ 3 and at the lower point y = − 1 − x2 , =√ = +1/ 3.
3/4 dx 1−x 2

√ √
28. If y 2 − x + 1 = 0, then y = x − 1 goes through the point (10, 3) so dy/dx = 1/(2 x − √ By1).
implicit diﬀerentiation dy/dx = 1/(2y). In both cases, dy/dx|(10,3) = 1/6. Similarly y = − x − 1
√
goes through (10, −3) so dy/dx = −1/(2 x − 1) = −1/6 which yields dy/dx = 1/(2y) = −1/6.

dy dy x3 1
29. 4x3 + 4y 3 = 0, so = − 3 = − 3/4 ≈ −0.1312.
dx dx y 15

dy dy dy dy y+1
30. 3y 2 + x2 + 2xy + 2x − 6y = 0, so = −2x 2 = 0 at x = 0
dx dx dx dx 3y + x2 − 6y

dy dy
31. 4(x2 + y 2 ) 2x + 2y = 25 2x − 2y ,
dx dx
dy x[25 − 4(x2 + y 2 )] dy
= ; at (3, 1) = −9/13
dx y[25 + 4(x2 + y 2 )] dx

2 dy dy y 1/3 √ √
32. x−1/3 + y −1/3 = 0, = − 1/3 = 3 at (−1, 3 3)
3 dx dx x

da da da da 2t3 + 3a2
33. 4a3 − 4t3 = 6 a2 + 2at , solve for to get = 3
dt dt dt dt 2a − 6at

√
1 −1/2 du 1 −1/2 du u
34. u + v = 0 so = −√
2 dv 2 dv v

dω dω b2 λ dx dx 1
35. 2a2 ω + 2b2 λ = 0 so =− 2 36. 1 = (cos x) so = = sec x
dλ dλ a ω dy dy cos x


130 Chapter 4

37. (a) y

2
x
–4 4

–2

dy dy
(b) Implicit differentiation of the equation of the curve yields (4y 3 + 2y) = 2x − 1 so =0
dx dx
only if x = 1/2 but y 4 + y 2 ≥ 0, so x = 1/2 is impossible.
1± 1 + 4y 2 + 4y 4
(c) x2 − x − (y 4 + y 2 ) = 0, so by the Quadratic Formula x = = 1 + y 2 , −y 2
2
which gives the parabolas x = 1 + y 2 , x = −y 2 .

38. (a) y
2

x
0 1 2

–2

dy dy
(b) 2y = (x − a)(x − b) + x(x − b) + x(x − a) = 3x2 − 2(a + b)x + ab. If = 0 then
dx dx
3x2 − 2(a + b)x + ab = 0. By the Quadratic Formula
2(a + b) ± 4(a + b)2 − 4 · 3ab 1
x= = a + b ± (a2 + b2 − ab)1/2 .
6 3
(c) y = ± x(x − a)(x − b). The square root is only defined for nonnegative arguments, so it is
necessary that all three of the factors x, x − a, x − b be nonnegative, or that two of them be
nonpositive. If, for example, 0 < a < b then the function is defined on the disjoint intervals
0 < x < a and b < x < +∞, so there are two parts.

39. (a) y (b) x ≈ ±1.1547
2

x
–2 2

–2

dy dy dy y − 2x dy
(c) Implicit differentiation yields 2x − x − y + 2y = 0. Solve for = . If =0
dx dx dx 2y − x dx
2
then y − 2x = 0 or y = 2x. Thus 4 = x2 − xy + y 2 = x2 − 2x2 + 4x2 = 3x2 , x = ± √ .
3

40. (a) See Exercise 39 (a)
(b) Since the equation is symmetric in x and y, we obtain, as in Exercise 39, x ≈ ±1.1547.



dy dy dx 2y − x dx
(c) Implicit differentiation yields 2x − x − y + 2y = 0. Solve for = . If =0
dx dx dy y − 2x dy
2 4
then 2y − x = 0 or x = 2y. Thus 4 = 4y 2 − 2y 2 + y 2 = 3y 2 , y = ± √ , x = 2y = ± √ .
3 3

41. Solve the simultaneous equations y = x, x2 −xy+y 2 = 4 to get x2 −x2 +x2 = 4, x = ±2, y = x = ±2,
so the points of intersection are (2, 2) and (−2, −2).
dy y − 2x dy dy
From Exercise 39 part (c), = . When x = y = 2, = −1; when x = y = −2, = −1;
dx 2y − x dx dx
the slopes are equal.

42. Suppose a2 − 2ab + b2 = 4. Then (−a)2 − 2(−a)(−b) + (−b)2 = a2 − 2ab + b2 = 4 so if P (a, b) lies
on C then so does Q(−a, −b).
dy y − 2x dy b − 2a
From Exercise 39 part (c), = . When x = a, y = b then = , and when
dx 2y − x dx 2b − a
dy b − 2a
x = −a, y = −b, then = , so the slopes at P and Q are equal.
dx 2b − a

43. The point (1,1) is on the graph, so 1 + a = b. The slope of the tangent line at (1,1) is −4/3; use
dy 2xy 2 4
implicit differentiation to get =− 2 so at (1,1), − = − , 1 + 2a = 3/2, a = 1/4
dx x + 2ay 1 + 2a 3
and hence b = 1 + 1/4 = 5/4.

44. The slope of the line x + 2y − 2 = 0 is m1 = −1/2, so the line perpendicular has slope m = 2
(negative reciprocal). The slope of the curve y 3 = 2x2 can be obtained by implicit differentiation:
dy dy 4x dy 4x
3y 2 = 4x, = 2 . Set = 2; 2 = 2, x = (3/2)y 2 . Use this in the equation of the curve:
dx dx 3y dx 3y
2
3 2 2
y 3 = 2x2 = 2((3/2)y 2 )2 = (9/2)y 4 , y = 2/9, x = = .
2 9 27

45. (a) y (b) x ≈ 0.84
2

x
–3 –1 2
–1

–3

(c) Use implicit differentiation to get dy/dx = (2y −3x2 )/(3y 2 −2x), so dy/dx = 0 if y = (3/2)x2 .
Substitute this into x3 − 2xy + y 3 = 0 to obtain 27x6 − 16x3 = 0, x3 = 16/27, x = 24/3 /3
and hence y = 25/3 /3.

46. (a) y (b) Evidently the tangent line at the point
2
x = 1, y = 1 has slope −1.

x
–3 –1 2
–1

–3


132 Chapter 4

(c) Use implicit differentiation to get dy/dx = (2y −3x2 )/(3y 2 −2x), so dy/dx = −1 if 2y −3x2 =
−3y 2 +2x, 2(y−x)+3(y−x)(y+x) = 0. One solution is y = x; this together with x3 +y 3 = 2xy
yields x = y = 1. For these values dy/dx = −1, so that (1, 1) is a solution.
To prove that there is no other solution, suppose y = x. From dy/dx = −1 it follows
that 2(y − x) + 3(y − x)(y + x) = 0. But y = x, so x + y = −2/3. Then x3 + y 3 =
(x + y)(x2 − xy + y 2 ) = 2xy, so replacing x + y with −2/3 we get x2 + 2xy + y 2 = 0, or
(x + y)2 = 0, so y = −x. Substitute that into x3 + y 3 = 2xy to obtain x3 − x3 = −2x2 , x = 0.
But at x = y = 0 the derivative is not defined.

47. (a) The curve is the circle (x − 2)2 + y 2 = 1 about the point (2, 0) of radius 1. One tangent
line is tangent at a point P(x,y) in the first quadrant. Let Q(2, 0) be the center of the
circle. Then OP Q is a right angle, with sides |P Q| = r = 1 and |OP | = x2 + y 2 . By
the Pythagorean Theorem x2 + y 2 + 12 = 22 . Substitute this into (x − 2)2 + y 2 = 1 to
√ √
obtain √ − 4x + 4 = 1, x = 3/2, y = 3 − x2 = 3/2. So the required tangent lines are
3
y = ±( 3/3)x.
(b) Let P (x0 , y0 ) be a point where a line through the origin is tangent to the curve
x2 − 4x + y 2 + 3 = 0. Implicit differentiation applied to the equation of the curve gives
dy/dx = (2 − x)/y. At P the slope of the curve must equal the slope of the line so
(2 − x0 )/y0 = y0 /x0 , or y0 = 2x0 − x2 . But x2 − 4x0 + y0 + 3 = 0 because (x0 , y0 ) is on the
2
0 0
2

curve, and elimination of y0 in the latter two equations gives x2 − 4x0 + (2x0 − x√) + 3 = 0,
2
0
2
0
x0 = 3/2 which when substituted into y0 =√ 0 − x0 yields y0 = 3/4, so y0 = ± 3/2. The
√
2
2x 2 2
√
slopes of the lines are (± 3/2)/(3/2) = ± 3/3 and their equations are y = ( 3/3)x and
√
y = −( 3/3)x.

48. Let P (x0 , y0 ) be a point where a line through the origin is tangent to the curve
2x2 − 4x + y 2 + 1 = 0. Implicit differentiation applied to the equation of the curve gives
dy/dx = (2 − 2x)/y. At P the slope of the curve must equal the slope of the line so
(2 − 2x0 )/y0 = y0 /x0 , or y0 = 2x0 (1 − x0 ). But 2x2 − 4x0 + y0 + 1 = 0 because (x0 , y0 ) is on the
2
0
2

curve, and elimination of y0 in the latter two equations gives 2x0 = 4x0 − 1, x0 = 1/2 which when
2
√
substituted into y0 = 2x0 (1 − x0 ) yields y0 = 1/2, √ y0 = ± 2/2. The slopes of the lines are
√
2
√
2
so √
(± 2/2)/(1/2) = ± 2 and their equations are y = 2x and y = − 2x.

49. The linear equation axr−1 x + by0 y = c is the equation of a line . Implicit differentiation of the
0
r−1

dy dy axr−1
equation of the curve yields raxr−1 + rby r−1 = 0, = − r−1 . At the point (x0 , y0 ) the slope
dx dx by
axr−1
of the line must be − r−1 , which is the slope of . Moreover, the equation of is satisfied by
0
by0
the point (x0 , y0 ), so this point lies on . By the point-slope formula, must be the line tangent
to the curve at (x0 , y0 ).
dy
50. Implicit differentiation of the equation of the curve yields rxr−1 + ry r−1 = 0. At the point (1, 1)
dy dy dx
this becomes r + r = 0, = −1.
dx dx
dy dy dt
51. By the chain rule, = . Use implicit differentiation on 2y 3 t + t3 y = 1 to get
dx dt dx
dy 2y 3 + 3t2 y dt 1 dy 2y 3 + 3t2 y
=− , but = so =− .
dt 6ty 2 + t3 dx cos t dx (6ty 2 + t3 ) cos t

4 1/3 4
52. (a) f (x) = x , f (x) = x−2/3
3 9
7 4/3 28 1/3 28 −2/3
(b) f (x) = x , f (x) = x , f (x) = x
3 9 27
(c) generalize parts (a) and (b) with k = (n − 1) + 1/3 = n − 2/3



53. y = rxr−1 , y = r(r − 1)xr−2 so 3x2 r(r − 1)xr−2 + 4x rxr−1 − 2xr = 0,
3r(r − 1)xr + 4rxr − 2xr = 0, (3r2 + r − 2)xr = 0,
3r2 + r − 2 = 0, (3r − 2)(r + 1) = 0; r = −1, 2/3

54. y = rxr−1 , y = r(r − 1)xr−2 so 16x2 r(r − 1)xr−2 + 24x rxr−1 + xr = 0,
16r(r − 1)xr + 24rxr + xr = 0, (16r2 + 8r + 1)xr = 0,
16r2 + 8r + 1 = 0, (4r + 1)2 = 0; r = −1/4

55. We shall find when the curves intersect and check that the slopes are negative reciprocals. For the
intersection solve the simultaneous equations x2 + (y − c)2 = c2 and (x − k)2 + y 2 = k 2 to obtain
1 y−c x−k
cy = kx = (x2 + y 2 ). Thus x2 + y 2 = cy + kx, or y 2 − cy = −x2 + kx, and =− .
2 x y
dy x dy x−k
Differentiating the two families yields (black) =− , and (gray) =− . But it was
dx y−c dx y
proven that these quantities are negative reciprocals of each other.

dy dy
56. Differentiating, we get the equations (black) x + y = 0 and (gray) 2x − 2y = 0. The first
dx dx
y x
says the (black) slope is = − and the second says the (gray) slope is , and these are negative
x y
reciprocals of each other.

EXERCISE SET 4.2

1 1 1 1 1
1. (5) = 2. =
5x x x/3 3 x

1 1 1 1
3. 4. √ √ = √ √
1+x 2+ x 2 x 2 x(2 + x)

1 2x 3x2 − 14x
5. 2−1
(2x) = 2 6.
x x −1 x3 − 7x2 − 3

1 (1 + x2 )(1) − x(2x) 1 − x2
7. 2) 2 )2
=
x/(1 + x (1 + x x(1 + x2 )

1 1−x+1+x 2 d d 2
8. = 9. (2 ln x) = 2 ln x =
(1 + x)/(1 − x) (1 − x)2 1 − x2 dx dx x

1 1 1 1
(ln x)−1/2
2
10. 3 (ln x) 11. = √
x 2 x 2x ln x

1 1 1 1
12. √ √ = 13. ln x + x = 1 + ln x
x2 x 2x x

1 2x2
14. x3 + (3x2 ) ln x = x2 (1 + 3 ln x) 15. 2x log2 (3 − 2x) −
x (3 − 2x) ln 2

3 2 2x − 2
16. log2 (x2 − 2x) + 3x log2 (x2 − 2x)
(x2 − 2x) ln 2


134 Chapter 4

2x(1 + log x) − x/(ln 10)
17. 18. 1/[x(ln 10)(1 + log x)2 ]
(1 + log x)2

1 1 1 1 1 1
19. = 20.
ln x x x ln x ln(ln(x)) ln x x

1 1
21. (sec2 x) = sec x csc x 22. (− sin x) = − tan x
tan x cos x

1 1 sin(2 ln x) sin(ln x2 )
23. − sin(ln x) 24. 2 sin(ln x) cos(ln x) = =
x x x x
1 cot x
25. 2 (2 sin x cos x) = 2 ln 10
ln 10 sin x
1 2 sin x cos x 2 tan x
26. (−2 sin x cos x) = − =−
(ln 10)(1 − sin x)
2 (ln 10) cos 2x ln 10

d 3 8x 11x2 − 8x + 3
27. 3 ln(x − 1) + 4 ln(x2 + 1) = + 2 =
dx x−1 x +1 (x − 1)(x2 + 1)

d 1 2x3
28. [2 ln cos x + ln(1 + x4 )] = −2 tan x +
dx 2 1 + x4

d 1 3x
29. ln cos x − ln(4 − 3x2 ) = − tan x +
dx 2 4 − 3x2

d 1 1 1 1
30. [ln(x − 1) − ln(x + 1)] = −
dx 2 2 x−1 x+1

1 dy 1 2x
31. ln |y| = ln |x| + ln |1 + x2 |,
3
= x 1 + x2 +
3 dx x 3(1 + x2 )

1 dy 1 x−1 1 1
32. ln |y| = [ln |x − 1| − ln |x + 1|], = 5
−
5 dx 5 x+1 x−1 x+1

1 1
33. ln |y| = ln |x2 − 8| + ln |x3 + 1| − ln |x6 − 7x + 5|
3 2
√
dy (x2 − 8)1/3 x3 + 1 2x 3x2 6x5 − 7
= + − 6
dx x6 − 7x + 5 3(x2 − 8) 2(x3 + 1) x − 7x + 5

1
34. ln |y| = ln | sin x| + ln | cos x| + 3 ln | tan x| − ln |x|
2
dy sin x cos x tan3 x 3 sec2 x 1
= √ cot x − tan x + −
dx x tan x 2x
√ √
√ 1 dy 10 dy 10
35. f (x) = ex e−1
36. ln y = − 10 ln x, =− , =− √
y dx x dx x1+ 10

ln e 1 d 1
37. (a) logx e = = , [logx e] = −
ln x ln x dx x(ln x)2
ln 2 d ln 2
(b) logx 2 = , [logx 2] = −
ln x dx x(ln x)2



ln b ln e 1
38. (a) From loga b = for a, b > 0 it follows that log(1/x) e = =− , hence
ln a ln(1/x) ln x
d 1
log(1/x) e =
dx x(ln x)2
ln e 1 d 1 1 1
(b) log(ln x) e = = , so log(ln x) e = − =−
ln(ln x) ln(ln x) dx (ln(ln x))2 x ln x x(ln x)(ln(ln x))2

1
39. f (x0 ) = f (e−1 ) = −1, f (x) = , f (x0 ) = e, y − (−1) = e(x − 1/e) = ex − 1, y = ex − 2
x

dy 1
40. y0 = log 10 = 1, y = log x = (log e) ln x, = log e ,
dx x=10 10
log e log e
y−1= (x − 10), y = x + 1 − log e
10 10

1 1 1
41. f (x0 ) = f (−e) = 1, f (x) =− , 42. y − ln 2 = − (x + 2), y = − x + ln 2 − 1
x=−e e 2 2
1 1
y − 1 = − (x + e), y = − x
e e

43. Let the equation of the tangent line be y = mx and suppose that it meets the curve at (x0 , y0 ).
1 1 1 ln x0 1
Then m = = and y0 = mx0 = ln x0 . So m = = and ln x0 = 1, x0 = e, m =
x x=x0 x0 x0 x0 e
1
and the equation of the tangent line is y = x.
e

44. Let y = mx + b be a line tangent to the curve at (x0 , y0 ). Then b is the y-intercept and the
1
slope of the tangent line is m = . Moreover, at the point of tangency, mx0 + b = ln x0 or
x0
1
x0 + b = ln x0 , b = ln x0 − 1, as required.
x0
y
45. The area of the triangle P QR, given by |P Q||QR|/2 is
required. |P Q| = w, and, by Exercise 44, |QR| = 1, so 1
P (w, ln w)
area = w/2. Q x
w 2
R

–2

46. Since y = 2 ln x, let y = 2z; then z = ln x and we apply the result of Exercise 45 to ﬁnd that the
area is, in the x-z plane, w/2. In the x-y plane, since y = 2z, the vertical dimension gets doubled,
so the area is w.

dy 1 dy 1
47. If x = 0 then y = ln e = 1, and = . But ey = x + e, so = y = e−y .
dx x+e dx e

dy 1
. But ey = e− ln(e −x) = (e2 − x)−1 , so
2
48. When x = 0, y = − ln(e2 ) = −2. Next, = 2
dx e −x
dy
= ey .
dx


136 Chapter 4

dy 1
49. Let y = ln(x + a). Following Exercise 47 we get = = e−y , and when x = 0, y = ln(a) = 0
dx x+a
if a = 1, so let a = 1, then y = ln(x + 1).

dy 1 1 dy
50. Let y = − ln(a − x), then = . But ey = , so = ey .
dx a−x a−x dx
If x = 0 then y = − ln(a) = − ln 2 provided a = 2, so y = − ln(2 − x).

ln(e2 + ∆x) − 2 d 1
51. (a) f (x) = ln x; f (e2 ) = lim = (ln x) = = e−2
∆x→0 ∆x dx x=e2 x x=e2

ln(1 + h) − ln 1 ln(1 + h) 1
(b) f (w) = ln w; f (1) = lim = lim = =1
h→0 h h→0 h w w=1

f (x) − f (0)
52. (a) Let f (x) = ln(cos x), then f (0) = ln(cos 0) = ln 1 = 0, so f (0) = lim =
x→0 x
ln(cos x)
lim , and f (0) = − tan 0 = 0.
x→0 x √
√
2 f (1 + h) − f (1) (1 + h) 2
−1
(b) Let f (x) = x , then f (1) = 1, so f (1) = lim = lim , and
√ √ √ h→0 h h→0 h
f (x) = 2x 2−1 , f (1) = 2.

d logb (x + h) − logb (x)
53. [logb x] = lim
dx h→0 h
1 x+h
= lim logb Theorem 1.6.2(b)
h→0 h x
1 h
= lim logb 1 +
h→0 h x
1
= lim logb (1 + v) Let v = h/x and note that v → 0 as h → 0
v→0 vx
1 1
= lim logb (1 + v) h and v are variable, whereas x is constant
x v→0 v
1
= lim log (1 + v)1/v Theorem 1.6.2.(c)
x v→0 b
1
= logb lim (1 + v)1/v Theorem 2.5.5
x v→0

1
= logb e Formula 7 of Section 7.1
x

EXERCISE SET 4.3
1. (a) f (x) = 5x4 + 3x2 + 1 ≥ 1 so f is one-to-one on −∞ < x < +∞.
d −1 1 1 1
(b) f (1) = 3 so 1 = f −1 (3); f (x) = −1 (x))
, (f −1 ) (3) = =
dx f (f f (1) 9

2. (a) f (x) = 3x2 + 2ex ; for −1 < x < 1, f (x) ≥ 2e−1 = 2/e, and for |x| > 1, f (x) ≥ 3x2 ≥ 3, so
f is increasing and one-to-one
d −1 1 1 1
(b) f (0) = 2 so 0 = f −1 (2); f (x) = , (f −1 ) (2) = =
dx f (f −1 (x)) f (0) 2



2 d −1 2
3. f −1 (x) = − 3, so directly f (x) = − 2 . Using Formula (1),
x dx x
−2 1
f (x) = , so = −(1/2)(f −1 (x) + 3)2 ,
(x + 3)2 f (f −1 (x))
2
d −1 2 2
f (x) = −(1/2) =−
dx x x2

ex − 1 d −1 ex 2
4. f −1 (x) = , so directly, f (x) = . Next, f (x) = , and using Formula (1),
2 dx 2 2x + 1
d −1 2f −1 (x) + 1 ex
f (x) = =
dx 2 2

5. (a) f (x) = 2x + 8; f < 0 on (−∞, −4) and f > 0 on (−4, +∞); not enough information. By
inspection, f (1) = 10 = f (−9), so not one-to-one
(b) f (x) = 10x4 + 3x2 + 3 ≥ 3 > 0; f (x) is positive for all x, so f is one-to-one
(c) f (x) = 2 + cos x ≥ 1 > 0 for all x, so f is one-to-one
x
(d) f (x) = −(ln 2) 1 < 0 because ln 2 > 0, so f is one-to-one for all x.
2

6. (a) f (x) = 3x2 + 6x = x(3x + 6) changes sign at x = −2, 0, so √ enough information; by
not √
observation (of the graph, and using some guesswork), f (−1 + 3) = −6 = f (−1 − 3), so
f is not one-to-one.
(b) f (x) = 5x4 + 24x2 + 2 ≥ 2 > 0; f is positive for all x, so f is one-to-one
1
(c) f (x) = ; f is one-to-one because:
(x + 1)2
if x1 < x2 < −1 then f > 0 on [x1 , x2 ], so f (x1 ) = f (x2 )
if −1 < x1 < x2 then f > 0 on [x1 , x2 ], so f (x1 ) = f (x2 )
if x1 < −1 < x2 then f (x1 ) > 1 > f (x2 ) since f (x) > 1 on (−∞, −1) and f (x) < 1 on
(−1, +∞)
d 1
(d) Note that f (x) is only deﬁned for x > 0. logb x = , which is always negative
dx x ln b
(0 < b < 1), so f is one-to-one.

dx dy 1
7. y = f −1 (x), x = f (y) = 5y 3 + y − 7, = 15y 2 + 1, = ;
dy dx 15y 2 + 1
dy dy dy 1
check: 1 = 15y 2 + , = 2+1
dx dx dx 15y

dx dy
8. y = f −1 (x), x = f (y) = 1/y 2 , = −2y −3 , = −y 3 /2;
dy dx
dy dy
check: 1 = −2y −3 , = −y 3 /2
dx dx

dx dy 1
9. y = f −1 (x), x = f (y) = 2y 5 + y 3 + 1, = 10y 4 + 3y 2 , = 4 + 3y 2
;
dy dx 10y
dy dy dy 1
check: 1 = 10y 4 + 3y 2 , =
dx dx dx 10y 4 + 3y 2

dx dy 1
10. y = f −1 (x), x = f (y) = 5y − sin 2y, = 5 − 2 cos 2y, = ;
dy dx 5 − 2 cos 2y
dy dy 1
check: 1 = (5 − 2 cos 2y) , =
dx dx 5 − 2 cos 2y


138 Chapter 4

12. −10xe−5x
2
11. 7e7x

1 1/x
13. x3 ex + 3x2 ex = x2 ex (x + 3) 14. − e
x2

dy (ex + e−x )(ex + e−x ) − (ex − e−x )(ex − e−x )
15. =
dx (ex + e−x )2

(e2x + 2 + e−2x ) − (e2x − 2 + e−2x )
= = 4/(ex + e−x )2
(ex + e−x )2

16. ex cos(ex )

dy (ln x)ex − ex (1/x) ex (x ln x − 1)
17. (x sec2 x + tan x)ex tan x 18. = =
dx (ln x)2 x(ln x)2

15 2
x (1 + 5x3 )−1/2 exp( 1 + 5x3 )
3x
19. (1 − 3e3x )e(x−e )
20.
2

(x − 1)e−x x−1 1
21. = x 22. [− sin(ex )]ex = −ex tan(ex )
1 − xe−x e −x cos(ex )

1
23. f (x) = 2x ln 2; y = 2x , ln y = x ln 2, y = ln 2, y = y ln 2 = 2x ln 2
y

1
24. f (x) = −3−x ln 3; y = 3−x , ln y = −x ln 3, y = − ln 3, y = −y ln 3 = −3−x ln 3
y

25. f (x) = π sin x (ln π) cos x;
1
y = π sin x , ln y = (sin x) ln π, y = (ln π) cos x, y = π sin x (ln π) cos x
y

26. f (x) = π x tan x (ln π)(x sec2 x + tan x);
1
y = π x tan x , ln y = (x tan x) ln π, y = (ln π)(x sec2 x + tan x)
y
y = π x tan x (ln π)(x sec2 x + tan x)

1 dy 3x2 − 2 1
27. ln y = (ln x) ln(x3 − 2x), = 3 ln x + ln(x3 − 2x),
y dx x − 2x x

dy 3x2 − 2 1
= (x3 − 2x)ln x 3 ln x + ln(x3 − 2x)
dx x − 2x x

1 dy sin x dy sin x
28. ln y = (sin x) ln x, = + (cos x) ln x, = xsin x + (cos x) ln x
y dx x dx x

1 dy 1
29. ln y = (tan x) ln(ln x), = tan x + (sec2 x) ln(ln x),
y dx x ln x
dy tan x
= (ln x)tan x + (sec2 x) ln(ln x)
dx x ln x



1 dy 2x 1
30. ln y = (ln x) ln(x2 + 3), = 2 ln x + ln(x2 + 3),
y dx x +3 x
dy 2x 1
= (x2 + 3)ln x 2 ln x + ln(x2 + 3)
dx x +3 x

31. f (x) = exe−1

32. (a) because xx is not of the form ax where a is constant
1
(b) y = xx , ln y = x ln x, y = 1 + ln x, y = xx (1 + ln x)
y

3 3 1/2 1
33. =√ 34. − =−
1− (3x)2 1 − 9x2 1− x+1 2 4 − (x + 1)2
2

1 1 sin x sin x 1, sin x > 0
35. (−1/x2 ) = − √ 36. √ = =
1 − 1/x2 |x| x2 − 1 1 − cos2 x | sin x| −1, sin x < 0

3x2 3x2 5x4 5
37. = 38. = √
1 + (x3 )2 1 + x6 |x5 | (x5 )2 − 1 |x| x10 − 1

39. y = 1/ tan x = cot x, dy/dx = − csc2 x

1
40. y = (tan−1 x)−1 , dy/dx = −(tan−1 x)−2
1 + x2

ex 1
41. √ + ex sec−1 x 42. − √
|x| x2 − 1 (cos−1 x) 1 − x2

3x2 (sin−1 x)2
43. 0 44. √ + 2x(sin−1 x)3
1 − x2

√
45. 0 46. −1/ e2x − 1

1 1 −1/2 1 1
47. − x =− √ 48. − √
1+x 2 2(1 + x) x −1
2 cot x(1 + x2 )

49. (a) Let x = f (y) = cot y, 0 < y < π, −∞ < x < +∞. Then f is diﬀerentiable and one-to-one
and f (f −1 (x)) = − csc2 (cot−1 x) = −x2 − 1 = 0, and
d 1 1
[cot−1 x] = lim = − lim 2 = −1.
dx x=0
x→0 f (f −1 (x)) x→0 x + 1

(b) If x = 0 then, from Exercise 50(a) of Section 1.5,
d d 1 1 1 1
cot−1 x = tan−1 = − 2 =− 2 . For x = 0, Part (a) shows the same;
dx dx x x 1 + (1/x)2 x +1
d 1
thus for −∞ < x < +∞, [cot−1 x] = − 2 .
dx x +1

d 1 du
(c) For −∞ < u < +∞, by the chain rule it follows that [cot−1 u] = − 2 .
dx u + 1 dx


140 Chapter 4

d d 1 1 1 −1
50. (a) By the chain rule, [csc−1 x] = sin−1 = − 2 = √
dx dx x x 1− (1/x)2 |x| x2 − 1
d du d −1 du
(b) By the chain rule, [csc−1 u] = [csc−1 u] = √
dx dx du |u| u2 − 1 dx

x (3x2 + tan−1 y)(1 + y 2 )
51. x3 + x tan−1 y = ey , 3x2 + y + tan−1 y = ey y , y =
1+y 2 (1 + y 2 )ey − x

1 1
52. sin−1 (xy) = cos−1 (x − y), (xy + y) = − (1 − y ),
1− x2 y 2 1 − (x − y)2
y 1 − (x − y)2 + 1 − x2 y 2
y =
1 − x2 y 2 − x 1 − (x − y)2

53. (a) f (x) = x3 − 3x2 + 2x = x(x − 1)(x − 2) so f (0) = f (1) = f (2) = 0 thus f is not one-to-one.
√
6 ± 36 − 24 √
(b) f (x) = 3x − 6x + 2, f (x) = 0 when x =
2
= 1 ± 3/3. f (x) > 0 (f is
√ 6 √ √
increasing) if x < 1 − 3/3, f (x) < 0 (f is decreasing) if 1 − 3/3 < x < 1 + 3/3, so f (x)
√ √ √
takes on values less than f (1 − 3/3) on both sides of 1 − 3/3 thus 1 − 3/3 is the largest
value of k.

54. (a) f (x) = x3 (x − 2) so f (0) = f (2) = 0 thus f is not one to one.
(b) f (x) = 4x3 − 6x2 = 4x2 (x − 3/2), f (x) = 0 when x = 0 or 3/2; f is decreasing on (−∞, 3/2]
and increasing on [3/2, +∞) so 3/2 is the smallest value of k.

55. (a) f (x) = 4x3 + 3x2 = (4x + 3)x2 = 0 only at x = 0. But on [0, 2], f has no sign change, so f
is one-to-one.
(b) F (x) = 2f (2g(x))g (x) so F (3) = 2f (2g(3))g (3). By inspection f (1) = 3, so
g(3) = f −1 (3) = 1 and g (3) = (f −1 ) (3) = 1/f (f −1 (3)) = 1/f (1) = 1/7 because
f (x) = 4x3 + 3x2 . Thus F (3) = 2f (2)(1/7) = 2(44)(1/7) = 88/7.
F (3) = f (2g(3)) = f (2·1) = f (2) = 24, so the line tangent to F (x) at (3, 25) has the equation
y − 25 = (88/7)(x − 3), y = (88/7)x − 89/7.

2 1
56. (a) f (x) = −e4−x 2+ < 0 for all x > 0, so f is one-to-one.
x2
(b) By inspection, f (2) = 1/2, so 2 = f −1 (1/2) = g(1/2). By inspection,
1 9
f (2) = − 2 + = − , and
4 4
d
F (1/2) = f ([g(x)]2 ) [g(x)2 ] = f ([g(x)]2 )2g(x)g (x)
dx x=1/2 x=1/2
−12 1
1 f (4) e (2 + 16 ) 33 11
= f (22 )2 · 2 =4 =4 = = 12
f (g(x)) x=1/2 f (2) (2 + 1 )
4
9e12 3e

57. (a) f (x) = kekx , f (x) = k 2 ekx , f (x) = k 3 ekx , . . . , f (n) (x) = k n ekx
(b) g (x) = −ke−kx , g (x) = k 2 e−kx , g (x) = −k 3 e−kx , . . . , g (n) (x) = (−1)n k n e−kx

dy
58. = e−λt (ωA cos ωt − ωB sin ωt) + (−λ)e−λt (A sin ωt + B cos ωt)
dt
= e−λt [(ωA − λB) cos ωt − (ωB + λA) sin ωt]



2 2
1 1 x−µ d 1 x−µ
59. f (x) = √ exp − −
2πσ 2 σ dx 2 σ
2
1 1 x−µ x−µ 1
=√ exp − −
2πσ 2 σ σ σ
2
1 1 x−µ
= −√ (x − µ) exp −
2πσ 3 2 σ

60. y = Aekt , dy/dt = kAekt = k(Aekt ) = ky

61. y = Ae2x + Be−4x , y = 2Ae2x − 4Be−4x , y = 4Ae2x + 16Be−4x so
y + 2y − 8y = (4Ae2x + 16Be−4x ) + 2(2Ae2x − 4Be−4x ) − 8(Ae2x + Be−4x ) = 0

62. (a) y = −xe−x + e−x = e−x (1 − x), xy = xe−x (1 − x) = y(1 − x)
(b) y = −x2 e−x + e−x = e−x (1 − x2 ), xy = xe−x
2 2 2 2
/2 /2 /2 /2
(1 − x2 ) = y(1 − x2 )

dy
63. = 100(−0.2)e−0.2x = −20y, k = −0.2
dx

64. ln y = (5x + 1) ln 3 − (x/2) ln 4, so
y /y = 5 ln 3 − (1/2) ln 4 = 5 ln 3 − ln 2, and
y = (5 ln 3 − ln 2)y

y 7e−t 7e−t + 5 − 5 1
65. ln y = ln 60 − ln(5 + 7e−t ), = = = 1 − y, so
y 5 + 7e−t 5 + 7e−t 12
dy y
=r 1− y, with r = 1, K = 12.
dt K

66. (a) 12

0 9
0

60 60 60
(b) P tends to 12 as t gets large; lim P (t) = lim = = = 12
t→+∞ t→+∞ 5 + 7e−t 5 + 7 lim e−t 5
t→+∞
(c) the rate of population growth tends to zero
3.2

0 9
0

10h − 1 d x d x ln 10
67. lim = 10 = e = ln 10
h→0 h dx x=0 dx x=0


142 Chapter 4

tan−1 (1 + h) − π/4 d 1 1
68. lim = tan−1 x = =
h→0 h dx x=1 1 + x2 x=1 2
√
9[sin−1 ( 23 + ∆x)]2 − π 2 d 3
69. lim = (3 sin−1 x)2 √
= 2(3 sin−1 x) √ √
∆x→0 ∆x dx x= 3/2 1 − x2 x= 3/2
π 3
= 2(3 ) = 12π
3 1 − (3/4)

(2 + ∆x)(2+∆x) − 4 d x d x ln x
70. lim = x = e
∆x→0 ∆x dx x=2 dx x=2

= (1 + ln x)ex ln x = (1 + ln 2)22 = 4(1 + ln 2)
x=2

√
3 sec−1 w − π d 3 3
71. lim = 3 sec−1 x = √ =
w→2 w−2 dx x=2 |2| 2 2−1 2

4(tan−1 w)w − π d d x ln tan−1 x
72. lim = 4(tan−1 x)x = 4e
w→1 w−1 dx x=1 dx x=1
2
1/(1 + x ) 14
= 4(tan−1 x)x ln tan−1 x + x = π ln(π/4) + == 2 + π ln(π/4)
tan−1 x x=1 2π

EXERCISE SET 4.4
x2 − 4 (x − 2)(x + 2) x+2 2
1. (a) lim = lim = lim =
x→2 x2 + 2x − 8 x→2 (x + 4)(x − 2) x→2 x + 4 3
5
2x − 5 2 − lim 2
x→+∞ x
(b) lim = =
x→+∞ 3x + 7 7 3
3 + lim
x→+∞ x

sin x cos x sin x
2. (a) = sin x = cos x so lim = lim cos x = 1
tan x sin x x→0 tan x x→0

x2 − 1 (x − 1)(x + 1) x+1 x2 − 1 2
(b) 3−1
= = 2 so lim 3 =
x (x − 1)(x 2 + x + 1) x +x+1 x→1 x − 1 3

π π
3. Tf (x) = −2(x + 1), Tg (x) = −3(x + 1), 4. Tf (x) = − x − , Tg (x) = − x −
2 2
limit = 2/3 limit = 1
x
e 1
5. lim =1 6. lim = 1/5
x→0 cos x x→3 6x − 13

sec2 θ tet + et
7. lim =1 8. lim = −1
θ→0 1 t→0 −et
cos x cos x
9. lim+ = −1 10. lim = +∞
x→π 1 x→0+ 2x

1/x 3e3x 9e3x
11. lim =0 12. lim = lim = +∞
x→+∞ 1 x→+∞ 2x x→+∞ 2



− csc2 x −x −1
13. lim+ = lim 2 = lim+ = −∞
x→0 1/x x→0 sin x
+ x→0 2 sin x cos x

−1/x x
14. lim = lim 1/x = 0
x→0+ (−1/x2 )e1/x x→0+ e

100x99 (100)(99)x98 (100)(99)(98) · · · (1)
15. lim x
= lim x
= · · · = lim =0
x→+∞ e x→+∞ e x→+∞ ex
√
cos x/ sin x 2/ 1 − 4x2
16. lim = lim+ cos2 x = 1 17. lim =2
x→0+ sec2 x/ tan x x→0 x→0 1

1
1− 1 1 x 1
18. lim 1 + x2 = lim = 19. lim xe−x = lim = lim x = 0
x→0 3x 2 x→0 3(1 + x2 ) 3 x→+∞ x→+∞ ex x→+∞ e

x−π 1
20. lim (x − π) tan(x/2) = lim = lim = −2
x→π x→π cot(x/2) x→π −(1/2) csc2 (x/2)

sin(π/x) (−π/x2 ) cos(π/x)
21. lim x sin(π/x) = lim = lim = lim π cos(π/x) = π
x→+∞ x→+∞ 1/x x→+∞ −1/x2 x→+∞

ln x 1/x − sin2 x −2 sin x cos x
22. lim tan x ln x = lim = lim+ = lim+ = lim+ =0
x→0+ x→0+ cot x x→0 − csc2 x x→0 x x→0 1

cos 5x −5 sin 5x −5(+1) 5
23. lim sec 3x cos 5x = lim = lim = =−
x→(π/2)− x→(π/2)− cos 3x x→(π/2)− −3 sin 3x (−3)(−1) 3

x−π 1
24. lim (x − π) cot x = lim = lim =1
x→π x→π tan x x→π sec2 x

ln(1 − 3/x) −3
25. y = (1 − 3/x)x , lim ln y = lim = lim = −3, lim y = e−3
x→+∞ x→+∞ 1/x x→+∞ 1 − 3/x x→+∞

3 ln(1 + 2x) 6
26. y = (1 + 2x)−3/x , lim ln y = lim − = lim − = −6, lim y = e−6
x→0 x→0 x x→0 1 + 2x x→0

ln(ex + x) ex + 1
27. y = (ex + x)1/x , lim ln y = lim = lim x = 2, lim y = e2
x→0 x→0 x x→0 e + x x→0

b ln(1 + a/x) ab
28. y = (1 + a/x)bx , lim ln y = lim = lim = ab, lim y = eab
x→+∞ x→+∞ 1/x x→+∞ 1 + a/x x→+∞

ln(2 − x) 2 sin2 (πx/2)
29. y = (2 − x)tan(πx/2) , lim ln y = lim = lim = 2/π, lim y = e2/π
x→1 x→1 cot(πx/2) x→1 π(2 − x) x→1

2 ln cos(2/x) (−2/x2 )(− tan(2/x))
30. y = [cos(2/x)]x , lim ln y = lim = lim
x→+∞ x→+∞ 1/x2 x→+∞ −2/x3
− tan(2/x) (2/x2 ) sec2 (2/x)
= lim = lim = −2, lim y = e−2
x→+∞ 1/x x→+∞ −1/x2 x→+∞

1 1 x − sin x 1 − cos x sin x
31. lim − = lim = lim = lim =0
x→0 sin x x x→0 x sin x x→0 x cos x + sin x x→0 2 cos x − x sin x


144 Chapter 4

1 − cos 3x 3 sin 3x 9 9
32. lim 2
= lim = lim cos 3x =
x→0 x x→0 2x x→0 2 2

(x2 + x) − x2 x 1
33. lim √ = lim √ = lim = 1/2
x→+∞ x2+x+x x→+∞ x2+x+x x→+∞ 1 + 1/x + 1

ex − 1 − x ex − 1 ex
34. lim = lim x = lim x = 1/2
x→0 xex − x x→0 xe + ex − 1 x→0 xe + 2ex

ex
35. lim [x − ln(x2 + 1)] = lim [ln ex − ln(x2 + 1)] = lim ln ,
x→+∞ x→+∞ x→+∞ x2 + 1
x x x
e e e
lim = lim = lim = +∞ so lim [x − ln(x2 + 1)] = +∞
x→+∞ x2 + 1 x→+∞ 2x x→+∞ 2 x→+∞

x 1
36. lim ln = lim ln = ln(1) = 0
x→+∞ 1 + x x→+∞ 1/x + 1

ln x 1/x 1
38. (a) lim = lim = lim =0
xn
x→+∞ x→+∞ nxn−1 x→+∞ nxn

xn nxn−1
(b) lim = lim = lim nxn = +∞
x→+∞ ln x x→+∞ 1/x x→+∞

3x2 − 2x + 1 0
39. (a) L’Hˆpital’s Rule does not apply to the problem lim
o because it is not a form.
x→1 3x2 − 2x 0
3x2 − 2x + 1
(b) lim =2
x→1 3x2 − 2x

e3x −12x+12
2
e0
40. L’Hˆpital’s Rule does not apply to the problem
o , which is of the form , and from
x4 − 16 0
which it follows that lim− and lim+ exist, with values −∞ if x approaches 2 from the left and
x→2 x→2
+∞ if from the right. The general limit lim does not exist.
x→2

1/(x ln x) 2
41. lim √ = lim √ =0 0.15
x→+∞ 1/(2 x) x→+∞ x ln x

100 10000
0

ln x
42. y = xx , lim ln y = lim = lim −x = 0, lim y = 1 1
x→0+ x→0+ 1/x x→0+ x→0+

0 0.5
0



43. y = (sin x)3/ ln x , 25
3 ln sin x x
lim ln y = lim = lim (3 cos x) = 3,
x→0+ x→0+ ln x x→0+ sin x
lim y = e3
x→0+

0 0.5
19

4 sec2 x 4 4.1
44. lim − = lim =4
x→π/2 sec x tan x x→π/2− sin x

1.4 1.6
3.3

1 e−x ln x − 1
45. ln x − ex = ln x − = ; 0
e−x e−x 0 3

ln x 1/x
lim e−x ln x = lim = lim = 0 by L’Hˆpital’s Rule,
o
x→+∞ x→+∞ ex x→+∞ ex

e−x ln x − 1
so lim [ln x − ex ] = lim = −∞
x→+∞ x→+∞ e−x
–16

ex
46. lim [ln ex − ln(1 + 2ex )] = lim ln –0.6
x→+∞ x→+∞ 1 + 2ex 0 12
1 1
= lim ln = ln ;
x→+∞ e−x +2 2
horizontal asymptote y = − ln 2

–1.2

1.02
47. y = (ln x)1/x ,
ln(ln x) 1
lim ln y = lim = lim = 0;
x→+∞ x→+∞ x x→+∞ x ln x

lim y = 1, y = 1 is the horizontal asymptote
x→+∞

100 10000
1


146 Chapter 4

x+1
x+1
x ln
48. y = , lim ln y = lim x+2 1
x+2 x→+∞ x→+∞ 1/x
−x2
= lim = −1;
x→+∞ (x + 1)(x + 2)

lim y = e−1 is the horizontal asymptote
x→+∞

0 50
0

49. (a) 0 (b) +∞ (c) 0 (d) −∞ (e) +∞ (f ) −∞

(ln a) ln x (ln a)/x
50. (a) Type 00 ; y = x(ln a)/(1+ln x) ; lim ln y = lim = lim+ = lim+ ln a = ln a,
x→0+ x→0+ 1 + ln x x→0 1/x x→0
lim y = eln a = a
x→0+

(b) Type ∞0 ; same calculation as Part (a) with x → +∞
(ln a) ln(x + 1) ln a
(c) Type 1∞ ; y = (x + 1)(ln a)/x , lim ln y = lim = lim = ln a,
x→0 x→0 x x→0 x + 1
lim y = eln a = a
x→0

1 + 2 cos 2x x + sin 2x sin 2x
51. lim does not exist, nor is it ±∞; lim = lim 1+ =1
x→+∞ 1 x→+∞ x x→+∞ x

2 − cos x 2x − sin x 2 − (sin x)/x 2
52. lim does not exist, nor is it ±∞; lim = lim =
x→+∞ 3 + cos x x→+∞ 3x + sin x x→+∞ 3 + (sin x)/x 3

x(2 + sin 2x) 2 + sin 2x
53. lim (2 + x cos 2x + sin 2x) does not exist, nor is it ±∞; lim = lim ,
x→+∞ x+1
x→+∞ x→+∞ 1 + 1/x
which does not exist because sin 2x oscillates between −1 and 1 as x → +∞

1 1 sin x
54. lim + cos x + does not exist, nor is it ±∞;
x→+∞ x 2 2x
x(2 + sin x) 2 + sin x
lim = lim =0
x→+∞ x2 + 1 x→+∞ x + 1/x

V t −Rt/L
L e Vt
55. lim+ =
R→0 1 L

π/2 − x −1
56. (a) lim (π/2 − x) tan x = lim = lim = lim sin2 x = 1
x→π/2 x→π/2 cot x x→π/2 − csc2 x x→π/2

1 1 sin x cos x − (π/2 − x) sin x
(b) lim − tan x = lim − = lim
x→π/2 π/2 − x x→π/2 π/2 − x cos x x→π/2 (π/2 − x) cos x
−(π/2 − x) cos x
= lim
x→π/2 −(π/2 − x) sin x − cos x
(π/2 − x) sin x + cos x
= lim =0
x→π/2 −(π/2 − x) cos x + 2 sin x

(c) 1/(π/2 − 1.57) ≈ 1255.765534, tan 1.57 ≈ 1255.765592;
1/(π/2 − 1.57) − tan 1.57 ≈ 0.000058


Review Exercises, Chapter 4 147

kt − 1 (ln k)k t
57. (b) lim x(k 1/x − 1) = lim = lim = ln k
x→+∞ t
t→0+ t→0 + 1
√
(c) ln 0.3 = −1.20397, 1024 1024 0.3 − 1 = −1.20327;
√
ln 2 = 0.69315, 1024 1024 2 − 1 = 0.69338

k + cos x
58. If k = −1 then lim (k + cos x) = k + 1 = 0, so lim = ±∞. Hence k = −1, and by the
x→0 x→0 x2
−1 + cos x − sin x − 2 cos x 2 √
rule lim = lim = lim = − = −4 if = ±2 2.
x→0 x2 x→0 2x x→0 2 2

59. (a) No; sin(1/x) oscillates as x → 0. (b) 0.05

–0.35 0.35

–0.05

(c) For the limit as x → 0 use the Squeezing Theorem together with the inequalities
+

−x2 ≤ x2 sin(1/x) ≤ x2 . For x → 0− do the same; thus lim f (x) = 0.
x→0

− cos(1/x) + 2x sin(1/x)
60. (a) Apply the rule to get lim which does not exist (nor is it ±∞).
x→0 cos x
x x 1
(b) Rewrite as lim [x sin(1/x)], but lim = lim = 1 and lim x sin(1/x) = 0,
x→0 sin x x→0 sin x x→0 cos x x→0
x
thus lim [x sin(1/x)] = (1)(0) = 0
x→0 sin x

sin(1/x) sin x
61. lim , lim+ = 1 but lim+ sin(1/x) does not exist because sin(1/x) oscillates between
x→0+ (sin x)/x x→0 x x→0

x sin(1/x)
−1 and 1 as x → +∞, so lim+ does not exist.
x→0 sin x

f (x) (f (x) − f (0)/(x − a)
62. Since f (0) = g(0) = 0, then for x = a, = . Now take the limit:
g(x) (g(x) − g(0))/(x − a)
f (x) (f (x) − f (0)/(x − a) f (a)
lim = lim =
x→a g(x) x→a (g(x) − g(0))/(x − a) g (a)

REVIEW EXERCISES, CHAPTER 4
1 3 1 2x + 1
1. (6) = 2. (2x + 1) =
4(6x − 5)3/4 2(6x − 5)3/4 3(x2 + x)2/3 3(x2 + x)2/3

1/2 1/2
3 x−1 d x−1 9 x−1
3. dy/dx = =
2 x+2 dx x + 2 2(x + 2)2 x + 2

4
x2 (3 − 2x)1/3 (−2) − (3 − 2x)4/3 (2x) 2(3 − 2x)1/3 (2x − 9)
4. dy/dx = 3 =
x4 3x3


148 Chapter 4

dy dy 2 − y − 3x2
5. (a) 3x2 + x + y − 2 = 0, =
dx dx x
(b) y = (1 + 2x − x3 )/x = 1/x + 2 − x2 , dy/dx = −1/x2 − 2x
dy 2 − (1/x + 2 − x2 ) − 3x2
(c) = = −1/x2 − 2x
dx x
dy dy dy 1−y
6. (a) xy = x − y, x +y =1− , =
dx dx dx x+1
x 1
(b) y(x + 1) = x, y = ,y =
x+1 (x + 1)2
dy 1−y 1 − x+1
x
1
(c) = = = 2
dx x+1 1+x x +1
1 dy 1 dy y2
7. − − 2 = 0 so =− 2
y 2 dx x dx x

dy dy dy dy x2 − 2y
8. 3x2 − 3y 2 = 6(x + y), −(3y 2 + 6x) = 6y − 3x2 so = 2
dx dx dx dx y + 2x

dy dy dy y sec(xy) tan(xy)
9. x + y sec(xy) tan(xy) = , =
dx dx dx 1 − x sec(xy) tan(xy)

(1 + csc y)(− csc2 y)(dy/dx) − (cot y)(− csc y cot y)(dy/dx)
10. 2x = ,
(1 + csc y)2
dy
2x(1 + csc y)2 = − csc y(csc y + csc2 y − cot2 y) ,
dx
dy 2x(1 + csc y)
but csc2 y − cot2 y = 1, so =−
dx csc y

dy 3x d2 y (4y)(3) − (3x)(4dy/dx) 12y − 12x(3x/(4y)) 12y 2 − 9x2 −3(3x2 − 4y 2 )
11. = , 2
= 2
= 2
= 3
= ,
dx 4y dx 16y 16y 16y 16y 3
d2 y −3(7) 21
but 3x2 − 4y 2 = 7 so = =−
dx2 16y 3 16y 3
dy y
12. = ,
dx y−x
y y
(y − x) −y −1
d2 y (y − x)(dy/dx) − y(dy/dx − 1) y−x y−x
= =
dx 2 (y − x)2 (y − x)2
y 2 − 2xy d2 y 3
= but y 2 − 2xy = −3, so =−
(y − x)3 dx2 (y − x)3

dy dy dy dy dy 2
13. = tan(πy/2) + x(π/2) sec2 (πy/2), = 1 + (π/4) (2), =
dx dx dx dx dx π−2
14. Let P (x0 , y0 ) be the required point. The slope of the line 4x − 3y + 1 = 0 is 4/3 so the slope of
the tangent to y 2 = 2x3 at P must be −3/4. By implicit diﬀerentiation dy/dx = 3x2 /y, so at P ,
3x2 /y0 = −3/4, or y0 = −4x2 . But y0 = 2x3 because P is on the curve y 2 = 2x3 . Elimination of
0 0
2
0
y0 gives 16x0 = 2x0 , x0 (8x0 − 1) = 0, so x0 = 0 or 1/8. From y0 = −4x2 it follows that y0 = 0
4 3 3
0
when x0 = 0, and y0 = −1/16 when x0 = 1/8. It does not follow, however, that (0, 0) is a solution
because dy/dx = 3x2 /y (the slope of the curve as determined by implicit diﬀerentiation) is valid
only if y = 0. Further analysis shows that the curve is tangent to the x-axis at (0, 0), so the point
(1/8, −1/16) is the only solution.



15. Substitute y = mx into x2 + xy + y 2 = 4 to get x2 + mx2 + m2 x2 = 4, which has distinct solutions
√
x = ±2/ m2 + m + 1. They are distinct because m2 + m + 1 = (m + 1/2)2 + 3/4 ≥ 3/4, so
m2 + m + 1 is never zero.
Note that the points of intersection occur in pairs (x0 , y0 ) and (−x0 , y0 ). By implicit differentiation,
the slope of the tangent line to the ellipse is given by dy/dx = −(2x + y)/(x + 2y). Since the slope
is unchanged if we replace (x, y) with (−x, −y), it follows that the slopes are equal at the two point
of intersection.
Finally we must examine the special case x = 0 which cannot be written in the form y = mx. If
x = 0 then y = ±2, and the formula for dy/dx gives dy/dx = −1/2, so the slopes are equal.

16. Use implicit differentiation to get dy/dx = (y − 3x2 )/(3y 2 − x), so dy/dx = 0 if y = 3x2 . Substitute
√ √
this into x3 − xy + y 3 = 0 to obtain 27x6 − 2x3 = 0, x3 = 2/27, x = 3 2/3 and hence y = 3 4/3.

17. By implicit differentiation, 3x2 − y − xy + 3y 2 y = 0, so y = (3x2 − y)/(x − 3y 2 ). This derivative
exists except when x = 3y 2 . Substituting this into the original equation x3 − xy + y 3 = 0, one has
27y 6 − 3y 3 + y 3 = 0, y 3 (27y 3 − 2) = 0. The unique solution in the first quadrant is
y = 21/3 /3, x = 3y 2 = 22/3 /3

18. By implicit differentiation, dy/dx = k/(2y) so the slope of the tangent to y 2 = kx at (x0 , y0 ) is
k
k/(2y0 ) if y0 = 0. The tangent line in this case is y − y0 = (x − x0 ), or 2y0 y − 2y0 = kx − kx0 .
2
2y0
But y0 = kx0 because (x0 , y0 ) is on the curve y 2 = kx, so the equation of the tangent line becomes
2

2y0 y − 2kx0 = kx − kx0 which gives y0 y = k(x + x0 )/2. If y0 = 0, then x0 = 0; the graph of
y 2 = kx has a vertical tangent at (0, 0) so its equation is x = 0, but y0 y = k(x + x0 )/2 gives the
same result when x0 = y0 = 0.

1 2 3 4
19. y = ln(x + 1) + 2 ln(x + 2) − 3 ln(x + 3) − 4 ln(x + 4), dy/dx = + − −
x+1 x+2 x+3 x+4

1 1
20. y = ln x + ln(x + 1) − ln sin x + ln cos x, so
2 3
dy 1 1 cos x sin x 5x + 3
= + − − = − cot x − tan x
dx 2x 3(x + 1) sin x cos x 6x(x + 1)

1 1 2 ln x
21. (2) = 1/x 22. 2(ln x) =
2x x x
1 1 1
23. 24. y = 3 ln(x + 1), y =
3x(ln x + 1)2/3 3(x + 1)

ln ln x 1
25. log10 ln x = ,y =
ln 10 (ln 10)(x ln x)

1 + ln x/ ln 10 ln 10 + ln x (ln 10 − ln x)/x + (ln 10 + ln x)/x 2 ln 10
26. y = = ,y = =
1 − ln x/ ln 10 ln 10 − ln x (ln 10 − ln x)2 x(ln 10 − ln x)2

3 1 3 2x3
27. y = ln x + ln(1 + x4 ), y = +
2 2 2x (1 + x4 )

1 1 sin x 2x 1 − 3x2
28. y = ln x + ln cos x − ln(1 + x2 ), y = − − = − tan x
2 2x cos x 1 + x2 2x(1 + x2 )

29. y = x2 + 1 so y = 2x.


150 Chapter 4

(1 + ex + e2x ) dy ex
30. y = ln = − ln(1 − ex ), =
(1 − ex )(1 + ex + e2x ) dx 1 − ex
√
x
√
x d √ √ √ √
31. y = 2e + 2xe x = 2e x + xe x
dx
abe−x 2
32. y = 33. y =
(1 + be−x )2 π(1 + 4x2 )
−1 ln 2 −1
34. y = e(sin x) ln 2
,y =√ 2sin x
1−x 2

y 1 dy x 1 x
−1 x
35. ln y = ex ln x, = ex + ln x , = xe ex + ln x = ex xe + xe ln x
y x dx x
ln(1 + x) y x/(1 + x) − ln(1 + x) 1 ln(1 + x)
36. ln y = , = = − ,
x y x2 x(1 + x) x2
dy 1 (1 + x)(1/x)
= (1 + x)(1/x)−1 − ln(1 + x)
dx x x2
2
37. y =
|2x + 1| (2x + 1)2 − 1
1 d 1 x
38. y = √ cos−1 x2 = − √ √
2 cos−1 x2 dx cos−1 x2 1 − x4
1 3 x 3x2 x4
39. ln y = 3 ln x − ln(x2 + 1), y /y = − 2 ,y=√ − 2
2 x x +1 x2 + 1 (x + 1)3/2

1 y 1 2x 2x 4x 4x x2 − 1
(ln(x2 −1)−ln(x2 +1)), −
3
40. ln y = = = so y =
3 y 3 x2 − 1 x2 + 1 3(x4− 1) 3(x4 − 1) x2 + 1
dy 1 1 dy dy
41. (b) y (c) = − so < 0 at x = 1 and > 0 at x = e
dx 2 x dx dx
6

4

2
x
1 2 3 4

(d) The slope is a continuous function which goes from a negative value at x = 1 to a positive
value at x = e; therefore it must take the value zero between, by the Intermediate Value
Theorem.
dy
(e) = 0 when x = 2
dx
dβ 10
42. β = 10 log I − 10 log I0 , =
dI I ln 10
dβ 1 dβ 1
(a) = db/W/m2 (b) = db/W/m2
dI I=10I0 I0 ln 10 dI I=100I0 10I0 ln 10
dβ 1
(c) = db/W/m2
dI I=100I0 100I0 ln 10

dy dx dy dy dx dx
43. Solve =3 given y = x ln x. Then = = (1 + ln x) , so 1 + ln x = 3, ln x = 2,
dt dt dt dx dt dt
x = e2 .



44. x = 2, y = 0; y = −2x/(5 − x2 ) = −4 at x = 2, so y − 0 = −4(x − 2) or y = −4x + 8

1 y
45. Set y = logb x and solve y = 1: y = = 1 so
x ln b
1
x = . The curves intersect when (x, x) lies on the
ln b 2
graph of y = logb x, so x = logb x. From Formula (8), x
ln x 2
Section 1.6, logb x = from which ln x = 1, x = e,
ln b
ln b = 1/e, b = e1/e ≈ 1.4447.

√ y
46. (a) Find the point of intersection: f (x) = x + k = ln x. The
2
1 1 √
slopes are equal, so m1 = = m2 = √ , x = 2, x = 4. x
x 2 x
√ 2
Then ln 4 = 4 + k, k = ln 4 − 2.

k 1
(b) Since the slopes are equal m1 = √ = m2 = , so y
2 x x
√ √ 2
k x = 2. At the point of intersection k x = ln x, 2 = ln x, x
x = e2 , k = 2/e. 0 5

47. Where f is differentiable and f = 0, g must be differentiable; this can be inferred from the graphs.
In general, however, g need not be differentiable: consider f (x) = x3 , g(x) = x1/3 .

48. (a) f (x) = −3/(x + 1)2 . If x = f (y) = 3/(y + 1) then y = f −1 (x) = (3/x) − 1, so
d −1 3 1 (f −1 (x) + 1)2 (3/x)2 3
f (x) = − 2 ; and −1 (x))
=− =− = − 2.
dx x f (f 3 3 x
d −1 2
(b) f (x) = ex/2 , f (x) = 1 ex/2 . If x = f (y) = ey/2 then y = f −1 (x) = 2 ln x, so
2 f (x) = ;
dx x
1 −1 2
and = 2e−f (x)/2 = 2e− ln x = 2x−1 =
f (f −1 (x)) x

49. Let P (x0 , y0 ) be a point on y = e3x then y0 = e3x0 . dy/dx = 3e3x so mtan = 3e3x0 at P and an
equation of the tangent line at P is y − y0 = 3e3x0 (x − x0 ), y − e3x0 = 3e3x0 (x − x0 ). If the line
passes through the origin then (0, 0) must satisfy the equation so −e3x0 = −3x0 e3x0 which gives
x0 = 1/3 and thus y0 = e. The point is (1/3, e).

dy/dx dy
50. ln y = ln 5000 + 1.07x; = 1.07, or = 1.07y
y dx

dy/dx dy
51. ln y = 2x ln 3 + 7x ln 5; = 2 ln 3 + 7 ln 5, or = (2 ln 3 + 7 ln 5)y
y dx

dk q(T − T0 ) q qk0 q(T − T0 )
52. = k0 exp − − =− exp −
dT 2T0 T 2T 2 2T 2 2T0 T

53. y = aeax sin bx + beax cos bx and y = (a2 − b2 )eax sin bx + 2abeax cos bx, so y − 2ay + (a2 + b2 )y
= (a2 − b2 )eax sin bx + 2abeax cos bx − 2a(aeax sin bx + beax cos bx) + (a2 + b2 )eax sin bx = 0.


152 Chapter 4

√ √ 1 −2x
54. sin(tan−1 x) = x/ 1 + x2 and cos(tan−1 x) = 1/ 1 + x2 , and y = ,y = , hence
1 + x2 (1 + x2 )2
−2x x 1
y + 2 sin y cos3 y = + 2√ = 0.
(1 + x2 )2 1+x2 (1 + x2 )3/2

55. (a) 100

0 8
20

(b) as t tends to +∞, the population tends to 19
95 95 95
lim P (t) = lim = = = 19
t→+∞ t→+∞ 5 − 4e−t/4 5 − 4 lim e−t/4 5
t→+∞

(c) the rate of population growth tends to zero 0
0 8

–80

(1 + h)π − 1 d
56. (a) y = (1 + x)π , lim = (1 + x)π = π(1 + x)π−1 =π
h→0 h dx x=0 x=0

1 − ln x 1 − ln x dy 1/x 1
(b) Let y = . Then y(e) = 0, and lim = =− =−
ln x x→e (x − e) ln x dx (ln x)2 e
x=e

57. In the case +∞ − (−∞) the limit is +∞; in the case −∞ − (+∞) the limit is −∞, because
large positive (negative) quantities are added to large positive (negative) quantities. The cases
+∞ − (+∞) and −∞ − (−∞) are indeterminate; large numbers of opposite sign are subtracted,
and more information about the sizes is needed.

58. (a) when the limit takes the form 0/0 or ∞/∞
(b) Not necessarily; only if lim f (x) = 0. Consider g(x) = x; lim g(x) = 0. For f (x) choose
x→a x→0
cos x x2 |x|1/2
cos x, x , and |x|
2 1/2
. Then: lim does not exist, lim = 0, and lim = +∞.
x→0 x x→0 x x→0 x2

ex ex ex
59. lim (ex − x2 ) = lim x2 (ex /x2 − 1), but lim = lim = lim = +∞
x→+∞ x→+∞ x→+∞ x2 x→+∞ 2x x→+∞ 2

so lim (ex /x2 − 1) = +∞ and thus lim x2 (ex /x2 − 1) = +∞
x→+∞ x→+∞

ln x 1/x 1 ln x ln x 1
60. lim = lim = ; lim = lim =
x→1 x4 − 1 x→1 4x 3 4 x→1 x4 −1 x→1 x4 −1 2

(x2 + 2x)ex (x2 + 2x)ex (x2 + 4x + 2)ex 1
61. = lim = lim = lim =
x→0 6 sin 3x cos 3x x→0 3 sin 6x x→0 18 cos 6x 9

62. lim ax ln a = ln a
x→0

Chapter 04

More Related Content

What's hot

Similar to Chapter 04

More from ramiz100111

Recently uploaded

Chapter 04